Griggs and Yeh's Conjecture and expansion

1. 1
2
For maximum degree of an L(2,1) graph, maximum number
of coloring will be ( 3)

   
We claim that for maximum degree  maximum coloring of a graph of radius 2 is more
than the maximum coloring of the graph of radius 1 i.e. complete graph. In the case of
complete graph K give the maximum color 2 among all complete graph.
But we can form a graph of radius 2 of maximum degree  and which give the color
more than 2 .
Since this graph is of radius 2 and has ( 1) 1 2 2        vertices then we can color
this by using minimum 2 1  colors, as in this kind of graph each vertex get different
colors. And in the previous graph we get that each vertex has maximum degree  and it
form the graph of radius 2.
Now in the case of L(2,1) and maximum degree  the maximum graph of radius 2 [the
graph containing maximum number of vertices having distance maximum 2 in between
them and the graph satisfying the condition that it has maximum degree  ] have highest
color than any other radius graphs of maximum degree  .
We can do it by specific formation of that graph by merging and ignoring some vertices
as needed

2. 2
Thus if we make a graph containing largest number of vertices (as largest as possible)
having distance maximum 2 in between them and maintain maximum degree  then the
coloring of this graph give maximum coloring
In this graph level 1 vertex is root, in level 2 the  number of neighbours and in level 3
 -1 neighbours of each neighbours in level 2, and suitable connection between them so
that the total graph can contain maximum number of vertices connected by at most
distance 2 in between them in such a way that the maximum degree  would maintained.
So all vertices in level 3 also have degree  . And this is the maximum graph among all
radius 2 graphs having maximum degree  .
Here total number of vertices be  (  -1) +  +1 = 2
1 
And the coloring of this kind of graph will be 2
 . And another possible graphs has less
vertices and less or equal colors.
So the maximum coloring will be 2
 
But we reduce this bound. If the maximum color is given by the
maximum subgraph of that specific maximum degree then we see that the maximum
color will be reduced to a function f(x) where f(3) = 9 and f(  ) = f(  -1) +  for   4.
We will make a special set of graphs kM 
in which all members are graph which is
minimum graph containing the most possible vertex in that graph joining by a distance
maximum k maintaining the maximum degree  . Then any graph of the kind
L(k,k-1,…1) of the maximum degree  can be converted to one of the graph of kM 
And for that the maximum color of L(k,k-1,…1) will come from the maximum colored
graph of kM 
. And using the previous formation for maximum degree  we have to go
upto level k (k>2). And we get the maximum color will be less than or equal to
1 2
( 1) ( 1) ... ( 1)k k 
              .